I think jogging/running/walking because you don't need any equipment and you can structure around it on your own time.
I would say true. If you are calculating using vectors than it would need both...
Answer:
0.0354 kg/s
Explanation:
= Initial displacement = 0.5 m
= Final displacement = 0.1 m
m = Mass of egg = 55 g
t = Time taken = 5 seconds
Displacement of the oscillator under damping motion is given by
![x=Ae^{-\frac{b}{2m}t}cos(\omega't+\phi)](https://tex.z-dn.net/?f=x%3DAe%5E%7B-%5Cfrac%7Bb%7D%7B2m%7Dt%7Dcos%28%5Comega%27t%2B%5Cphi%29)
For maximum displacement
![cos(\omega't+\phi)=1](https://tex.z-dn.net/?f=cos%28%5Comega%27t%2B%5Cphi%29%3D1)
![A_2=A_1e^{-\frac{b}{2m}t}](https://tex.z-dn.net/?f=A_2%3DA_1e%5E%7B-%5Cfrac%7Bb%7D%7B2m%7Dt%7D)
From this equation we get
![b=\frac{2m}{t}ln\frac{A_1}{A_2}\\\Rightarrow b=\frac{2\times 0.055}{5}ln\frac{0.5}{0.1}\\\Rightarrow b=0.0354\ kg/s](https://tex.z-dn.net/?f=b%3D%5Cfrac%7B2m%7D%7Bt%7Dln%5Cfrac%7BA_1%7D%7BA_2%7D%5C%5C%5CRightarrow%20b%3D%5Cfrac%7B2%5Ctimes%200.055%7D%7B5%7Dln%5Cfrac%7B0.5%7D%7B0.1%7D%5C%5C%5CRightarrow%20b%3D0.0354%5C%20kg%2Fs)
The magnitude of the damping constant is 0.0354 kg/s
Answer:
![v = 4.44 \times 10^5 m/s](https://tex.z-dn.net/?f=v%20%3D%204.44%20%5Ctimes%2010%5E5%20m%2Fs)
Explanation:
By Einstein's Equation of photoelectric effect we know that
![h\nu = W + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=h%5Cnu%20%3D%20W%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
here we know that
= energy of the photons incident on the metal
= minimum energy required to remove photons from metal
= kinetic energy of the electrons ejected out of the plate
now we know that it requires 351 nm wavelength of photons to just eject out the electrons
so we can say
![W = \frac{hc}{351 nm}](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7Bhc%7D%7B351%20nm%7D)
here we know that
![hc = 1242 eV-nm](https://tex.z-dn.net/?f=hc%20%3D%201242%20eV-nm)
now we have
![W = \frac{1242}{351} = 3.54 eV](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1242%7D%7B351%7D%20%3D%203.54%20eV)
now by energy equation above when photon of 303 nm incident on the surface
![\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2](https://tex.z-dn.net/?f=%5Cfrac%7B1242%20eV-nm%7D%7B303%20nm%7D%20%3D%203.54%20eV%20%2B%20%5Cfrac%7B1%7D%7B2%7D%289.1%20%5Ctimes%2010%5E%7B-31%7D%29v%5E2)
![4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2](https://tex.z-dn.net/?f=4.1%20eV%20%3D%203.54%20eV%20%2B%20%284.55%20%5Ctimes%2010%5E%7B-31%7D%29%20v%5E2)
![(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2](https://tex.z-dn.net/?f=%284.1%20-%203.54%29%5Ctimes%201.6%20%5Ctimes%2010%5E%7B-19%7D%29%20%3D%20%284.55%20%5Ctimes%2010%5E%7B-31%7D%29%20v%5E2)
![8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2](https://tex.z-dn.net/?f=8.96%20%5Ctimes%2010%5E%7B-20%7D%20%3D%20%284.55%20%5Ctimes%2010%5E%7B-31%7D%29%20v%5E2)
![v = 4.44 \times 10^5 m/s](https://tex.z-dn.net/?f=v%20%3D%204.44%20%5Ctimes%2010%5E5%20m%2Fs)
Answer:
Wow I hope to never have to learn this
Explanation: